Wave Pulse

A wave pulse travels down a slinky. The mass of the slinky is m = 0.86 kg and is initially stretched to a length L = 6.5 m. The wave pulse has an amplitude of A = 0.22 m and takes t = 0.494 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.48 Hz.

1.What is the average speed of a piece of the slinky as a complete wave pulse passes?

2.Now the slinky is stretched to twice its length (but the total mass does not change).
What is the new tension in the slinky? (assume the slinky acts as a spring that obeys Hooke’s Law)

2. Relevant equations
v=λ*f
ω=2*∏*f
v=√(F/μ)

3. The attempt at a solution

1. Tried to use v=λ*f but can't seem to find k for λ

2. Tried to use v=√(F/μ) to combine with hooke's law F=-kx but could not find the answer.